The cycloid

Around 1628 the Paris monk Marin Mersenne put to his friends a question that looked innocent. When a wheel rolls along a road, a nail driven into its rim traces arches in the air. How large is the area under one such arch? And what kind of curve is it, anyway?

Mersenne was the hub through which the correspondence of scientific Europe flowed. You can explore his network of correspondents on the map in Chapter VIII. The question about the wheel ran along this network in every direction, and thirty years later it had become one of the most famous disputes of the 17th century. That is why the curve came to be called the Helen of geometers: like the beautiful Helen of Troy, it set the best of them against each other.

§Who quarrelled

• Marin Mersenne1588–1648 · organiser of the dispute
• Gilles P. de Roberval1602–1675 · area and tangent
• René Descartes1596–1650 · tangent
• Pierre de Fermat1601–1665 · tangent, length
• Christopher Wren1632–1723 · length
• Blaise Pascal1623–1662 · chronicler and challenger

§Questions of the 17th century

Click a question to jump straight to the answer.

1. What is the roulette?
2. If a curve is given geometrically, how do we find its equations?
3. What is the area bounded by this curve?
4. What is the tangent to this curve at a given point, and how do we find it? (questions 4 and 5)
6. What is the length of this curve?
7. What can be said about the centre of gravity of the figure bounded by this curve?

Question 1What is the roulette?

In French roue means wheel and the suffix -ette marks something small, so roulette is a “little wheel”. Pascal wrote that this curve “is nothing other than the path traced by a nail on a wheel when the wheel turns in the ordinary way”.

In Greek kyklos is a circle and eidos a shape. The cycloid is the path of a point on the rim of a circle that rolls along a straight line without slipping.

§The family of roulettes

In general, a roulette is any curve traced by a point fixed to a curve (most often a circle) that rolls without slipping along another curve. The cycloid is just the simplest member of a large family. If the point is farther from the centre than the radius, it traces a prolate cycloid with loops. If it is closer, it traces a curtate cycloid, a wave without cusps.

Question 2If a curve is given geometrically, how do we find its equations?

When a circle of radius R turns through an angle θ, its centre moves forward by exactly the length of the arc that has rolled out, that is Rθ. The point on the rim is turned about the centre by the same angle. Adding the two motions gives:

x(θ) = R(θ − sin θ),  y(θ) = R(1 − cos θ).

Roberval saw this as a composition of two motions: the uniform forward motion of the centre (green) and the rotation, whose vertical projection draws a sine curve (blue). Their sum is the cycloid.

Question 3What is the area bounded by this curve?

1The companion curve

From a point P on the cycloid draw a horizontal line p, and from the centre of the circle a vertical line q. As the circle rolls, their intersection Q draws the companion of the cycloid, the sine curve y = R(1 − cos x/R). Both curves are linked by the same motion.

2The moustache and the hat

The companion splits the area under one arch of the cycloid into two parts: the “hat” under the sine curve and the “moustache”, the two teardrops between the sine curve and the cycloid. Divide the 2πR × 2R rectangle vertically into two halves. The sine curve is centrally symmetric about the centre of each of them, so it cuts each half in two. The hat therefore takes up exactly half of the rectangle:

hat = 12 · 2πR · 2R = 2πR² = the area of two circles.

An even more vivid trick: the hat can be cut and rearranged into a triangle. The sine curve passes through the midpoints of the sides of the triangle with base 2πR and height 2R, and it is centrally symmetric about them. The piece of the hat that sticks out above a side fits exactly into the gap by the base after a 180° turn about the midpoint of that side.

3The method of indivisibles

At every height y, the horizontal segment across a teardrop is exactly as long as the half-chord of the rolling circle, namely R sin θ. By Cavalieri’s principle the teardrops therefore have the same area as a semicircle: together the two teardrops make one circle.

The result. Hat + moustache = 2πR² + πR². The area under one arch of the cycloid is 3πR², exactly three times the rolling circle. Galileo with his scales came very close; he just did not believe the ratio could be exactly 3. Roberval gave the curve the name trochoid.

Today a single integral is enough:

∫₀^2πR y dx = R² ∫₀^2π (1 − cos θ)² dθ = 3πR².

Roberval managed it without integral calculus, though, and proudly wrote the result to Descartes.

Questions 4 and 5What is the tangent to this curve at a given point? How do we find it?

Roberval understood the tangent as the direction of instantaneous motion. A point of the cycloid moves in two ways at once: the horizontal motion of the centre (green vector) and the rotation around the circle (orange vector, tangent to the circle). When the wheel rolls without slipping, both have the same size. The diagonal of the parallelogram they form is the direction of the tangent.

One by one Fermat, Beaugrand and Galileo’s pupil Torricelli joined the dispute. When Torricelli published the computation of the area in 1644, a scandal over priority broke out: Roberval accused him of taking the result from the Paris letters. You can also try the construction in GeoGebra: Descartes’ tangent ↗.

Question 6What is the length of this curve?

The idea: instead of a circle we roll a regular n-gon. Its vertex then traces a curve made of circular arcs. Each arc is a rotation about one vertex through the angle 2π/n, with radius equal to the chord 2 sin(kπ/n) (for R = 1). The length of an arc is radius times angle, so

L_n = n−1Σk=1 2 sin kπn · 2πn.

Using Lagrange’s identity for a sum of sines we get L_n = 4πn · cot π2n, and in the limit

L_∞ = lim_n→∞ 8 · π/2ntan(π/2n) = 8.

The length of one arch of the cycloid is 8R, four times the diameter of the rolling circle. And it is rational, with no π! Wren was the first to state the result; Fermat and Roberval also gave proofs.